What is the general solution for 1+tan^2(θ)=4tan^2(θ) ?

The general solution for 1+tan^2(θ)=4tan^2(θ) is θ=0.52359…+pin,θ=-0.52359…+pin

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1+tan^2(θ)=4tan^2(θ)

1 + tan^{2} (θ) = 4 tan^{2} (θ)

θ = 0.52359 \dots + πn, θ = - 0.52359 \dots + πn

Solution steps

1 + tan^{2} (θ) = 4 tan^{2} (θ)

Solve by substitution

tan (θ) = \frac{1}{3}, tan (θ) = - \frac{1}{3}

tan (θ) = \frac{1}{3} : θ = arctan (\frac{1}{3}) + πn

tan (θ) = - \frac{1}{3} : θ = arctan (- \frac{1}{3}) + πn

Combine all the solutions

θ = arctan (\frac{1}{3}) + πn, θ = arctan (- \frac{1}{3}) + πn

Show solutions in decimal form

θ = 0.52359 \dots + πn, θ = - 0.52359 \dots + πn

(tan (θ) + 1) (cos (θ) + 1) (sin (θ)) = 0

6 sin (2 x) - 3 = 0

4 cos^{2} (x) - sin (x) = 1

0.6 = cos (4 x)

sin (x) = \frac{1}{1.33}

What is the general solution for 1+tan^2(θ)=4tan^2(θ) ?
The general solution for 1+tan^2(θ)=4tan^2(θ) is θ=0.52359…+pin,θ=-0.52359…+pin